Kinetic Theory Test

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Kinetic Theory Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Equation of gas in terms of pressure $$p$$ absolute temperature $$T$$ and density $$d$$ is :

A
$$\dfrac{p_{1}}{d_{1}T_{1}}=\dfrac{p_{2}}{d_{2}T_{2}}$$

B
$$\dfrac{p_{1}T_{1}}{d_{1}}=\dfrac{p_{2}T_{2}}{d_{2}}$$

C
$$\dfrac{p_{1}d_{2}}{T_{2}}=\dfrac{p_{2}d_{1}}{T_{1}}$$

D
$$\dfrac{p_{1}d_{1}}{T_{1}}=\dfrac{p_{2}d_{2}}{T_{2}}$$

Solution

We know that the density of an ideal gas is given by the expression,

$$d = \dfrac{PM}{RT}$$ , where M is the molar mass of the gas

now, this means that $$\dfrac{P}{dT} = \dfrac{R}{M}$$

Now, both R and M are constant for a particular gas and do not depend on the physical parameters of the gas

So, $$\dfrac{P_{1}}{d_{1}T_{1}} = \dfrac{p_{2}}{d_{2}T_{2}}$$

So, A is the correct answer.
Question 2
1 / -0

The density of air at N.T.P. is $$1.293 gm/lit$$. If the pressure is tripled keeping its temperature constant its density becomes

A
$$3.87 gm/ltr$$

B
$$1.293gm/ltr$$

C
$$2.586 gm/ltr$$

D
$$0.431 gm/ltr$$

Solution

First we derive an expression for density of a gas.
Density = $$\dfrac{mass(w)}{volume(V)}$$
but we know that PV = nRT
now n = $$\dfrac{w}{M}$$ where M is molecular weight
So, PV = $$\dfrac{w}{M}RT$$
this gives $$\dfrac{w}{V} = \dfrac{PM}{RT}$$ = density = d
Clearly at a constant temperature,
$$\dfrac{d_{1}}{P_{1}} = \dfrac{d_{2}}{P_{2}}$$
so since pressure is tripled, density also triples
new density = old density x 3
$$= 1.293 \times 3$$
$$ = 3.87 gm/ltr$$
Question 3
1 / -0

A gas is kept at 13$$^{0}$$C in a vessel. If the volume of the gas is kept constant and is heated, the pressure will be doubled to its initial pressure at a temperature

A
572 K

B
286 K

C
143 K

D
73 K

Solution

PV = nRT
since n and R and V are constant

$$P_{1}T_{2} = P_{2}T_{1}$$
here $$V_{1} = V_{2}$$
and $$T_{1} = 286$$
and $$P_{2} = 2P_{1}$$
and $$T_{2} = ?$$ (to be determined)
Putting all the values, we get $$T_{2}= \dfrac{P_2 T_1}{P_1}= \dfrac{2 P_1 \times 286}{P_1}= 572K$$
Question 4
1 / -0

A bubble rises from the bottom of a lake, 90m deep. On reaching the surface, its volume becomes(take atmospheric pressure equals to 10 m of water )

A
4 times

B
8 times

C
10 times

D
3 times

Solution

At the bottom, pressure = 90m + 10m (atmospheric pressure) = 100m of water
at the top, pressure = 10m (only atmospheric pressure) = 10m of water
Since pressure has become one-tenth, volume becomes 10 times (Boyle's law)
Question 5
1 / -0

A cylinder contains a gas at temperature of 27$$^{0}$$C and a pressure 1MPa. If the temperature of the gas is lowered to -23$$^{0}$$ C , the change in pressure is

A
1MPa

B
5/6MPa

C
1/6MPa

D
5MPa

Solution

PV = nRT
Since n and R are constant, we get
$$\frac{PV}{T}$$ = constant
so, $$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$P_{1} = 1MPa$$
$$T_{1} = 300K$$
$$P_{2} = ?$$
$$V_{2} = V_{1}$$
$$T_{2} = 250K$$
Putting all the values, we get : $$P_{2} = \frac{5}{6}MPa$$
So change in pressure $$\frac{1}{6}$$MPa
Question 6
1 / -0

A volume 'V' and temperature 'T' was obtained, as shown in the diagram, when a given mass of gas was heated. During the heating process, the pressure is

A
increased

B
decreased

C
remains constant

D
changed erratically

Solution

$$\frac {dV}{dT}= constant$$ from the plot.
But, $$\frac {dV}{dT}= T + \frac {dP}{dT}$$
$$ \therefore \frac {dP}{dT}=Constant -T$$
$$\therefore$$ P decreases as T increases.
Question 7
1 / -0

A gas at 627$$^{0}$$ C is cooled so that its pressure becomes $$\dfrac{1}{3}$$ of its initial value at constant volume. Its final temperature is

A
900 K

B
600 K

C
300 K

D
100K

Solution

PV = nRT
since n and R and V are constant

$$P_{1}T_{2} = P_{2}T_{1}$$
here $$V_{1} = V_{2}$$
and $$T_{1} = 900$$
and $$P_{2} = \frac{P_{1}}{3}$$
and $$T_{2} = ?$$ (to be determined)
Putting all the values, we get $$T_{2}$$ = 300K
Question 8
1 / -0

At 27$$^{0}$$ C certain gas occupied a volume of 4 litres. If the volume of this is to be increased to 12 litres at the same pressure, the gas is to be heated to a temperature

A
300 K

B
900 K

C
600 K

D
1200K

Solution

PV = nRT
since n and R are constant, we get
$$\frac{PV}{T}$$ = constant
so, $$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$V_{1} = 4L$$
$$T_{1} = 300K$$
$$P_{2} = P_{1}$$
$$V_{2} = 12L$$
$$T_{2} = ? $$
putting all the values, we get :
$$T_{2}$$ = 900K
Question 9
1 / -0

A P-V diagram is obtained by changing the temperature of the gas as shown. During this process the gas is :

A
heated continuously

B
cooled continuously

C
heated in the beginning but cooled towards the end

D
cooled in the beginning but heated towards the end

Solution

From A to B
First initially pressure decrease and volume increases
equation of curve is $$P = -kV + c$$
Therefore, from ideal gas equation
$$T=PV/nR$$
$$T=(-kV+c)V/nR=(-kV^{2}+cV)/nR$$
Therefore its a parabola therefore in this process from A to B volume is increased therefore, its a parabola vertically downwards if we take V on x-axis and T on y-axis
Therefore as volume increases temperature rises becomes maximum and then again starts decreasing
Hence option(C)
Question 10
1 / -0

For an isochoric process the temperature at which the pressure of a gas will be double that of its pressure at 270$$^{0}$$ C is

A
543$$^{0}$$ C

B
813$$^{0}$$ C

C
3$$^{0}$$ C

D
270$$^{0}$$ C

Solution

PV = nRT
Since n and R are constant, we get
$$\frac{PV}{T}$$ = constant
so, $$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$T_{1} = 546K$$
$$P_{2} = 2P_{1}$$
$$V_{2} = V_{1}$$
$$T_{2} = ?$$
Putting all the values, we get : $$T_{2}$$ = 537$$^{\circ}$$ = 813K

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Kinetic Theory Test - 24

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Kinetic Theory Test - 24

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Question 1

$$\dfrac{p_{1}}{d_{1}T_{1}}=\dfrac{p_{2}}{d_{2}T_{2}}$$

$$\dfrac{p_{1}T_{1}}{d_{1}}=\dfrac{p_{2}T_{2}}{d_{2}}$$

$$\dfrac{p_{1}d_{2}}{T_{2}}=\dfrac{p_{2}d_{1}}{T_{1}}$$

$$\dfrac{p_{1}d_{1}}{T_{1}}=\dfrac{p_{2}d_{2}}{T_{2}}$$

Solution

Question 2

$$3.87 gm/ltr$$

$$1.293gm/ltr$$

$$2.586 gm/ltr$$

$$0.431 gm/ltr$$

Solution

Question 3

572 K

286 K

143 K

73 K

Solution

Question 4

4 times

8 times

10 times

3 times

Solution

Question 5

1MPa

5/6MPa

1/6MPa

5MPa

Solution

Question 6

increased

decreased

remains constant

changed erratically

Solution

Question 7

900 K

600 K

300 K

100K

Solution

Question 8

300 K

900 K

600 K

1200K

Solution

Question 9

heated continuously

cooled continuously

heated in the beginning but cooled towards the end

cooled in the beginning but heated towards the end

Solution

Question 10

543$$^{0}$$ C

813$$^{0}$$ C

3$$^{0}$$ C

270$$^{0}$$ C

Solution

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